What are the steps to figure out this math word problem? 5.9% of women were diagnosed with diabetes, how many were not diagnosed
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<h3>Answer: x = 7 and y = 3</h3>
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Explanation:
Apply the difference of squares rule
x² - 4y² = 13
x² - (2y)² = 13
(x - 2y)(x + 2y) = 13
Since x and y are positive integers, this means x-2y and x+2y are both integers as well.
The value 13 is prime. Its only factors are 1 and 13
Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:
- A) x-2y = 1 and x+2y = 13
- B) x-2y = 13 and x+2y = 1
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Let's consider case A
We have this system of equations
Add the equations straight down
- x+x becomes 2x
- -2y+2y becomes 0y = 0 which goes away
- 1+13 becomes 14
Therefore we have 2x = 14 solve to x = 7
From here, plug this into either equation to solve for y
x-2y = 1
7 - 2y = 1
-2y = 1-7
-2y = -6
y = -6/(-2)
y = 3
You should get the same result if you used x+2y = 13
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Since we've found that x = 7 and y = 3, notice how case B is not possible
Example: x-2y = 13 becomes 7-2(3) = 13 which is false.
Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.
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Let's check those x and y values in the original equation
x² - 4y² = 13
7² - 4*(3)² = 13
49 - 4(9) = 13
49 - 36 = 13
13 = 13
The answer is confirmed.
<span>Ayesha's right. There's a good trick for knowing if a number is a multiple of nine called "casting out nines." We just add up the digits, then add up the digits of the sum, and so on. If the result is nine the original number is a multiple of nine. We can stop early if we recognize if a number along the way is or isn't a multiple of nine. The same trick works with multiples of three; we have one if we end with 3, 6 or 9.
So </span>
<span>has a sum of digits 31 whose sum of digits is 4, so this isn't a multiple of nine. It will give a remainder of 4 when divided by 9; let's check.
</span>
<span>
</span>Let's focus on remainders when we divide by nine. The digit summing works because 1 and 10 have the same remainder when divided by nine, namely 1. So we see multiplying by 10 doesn't change the remainder. So
has the same remainder as 
.
When Ayesha reverses the digits she doesn't change the sum of the digits, so she doesn't change the remainder. Since the two numbers have the same remainder, when we subtract them we'll get a number whose remainder is the difference, namely zero. That's why her method works.
<span>
It doesn't matter if the digits are larger or smaller or how many there are. We might want the first number bigger than the second so we get a positive difference, but even that doesn't matter; a negative difference will still be a multiple of nine. Let's pick a random number, reverse its digits, subtract, and check it's a multiple of nine:
</span>
So </span>
</span>
</span>Let's focus on remainders when we divide by nine. The digit summing works because 1 and 10 have the same remainder when divided by nine, namely 1. So we see multiplying by 10 doesn't change the remainder. So
When Ayesha reverses the digits she doesn't change the sum of the digits, so she doesn't change the remainder. Since the two numbers have the same remainder, when we subtract them we'll get a number whose remainder is the difference, namely zero. That's why her method works.
<span>
It doesn't matter if the digits are larger or smaller or how many there are. We might want the first number bigger than the second so we get a positive difference, but even that doesn't matter; a negative difference will still be a multiple of nine. Let's pick a random number, reverse its digits, subtract, and check it's a multiple of nine:
</span>